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Final

# SOS-_Final_Package_F11.pdf

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School
University of Waterloo
Department
Statistics
Course
STAT 230
Professor
phdstudent
Semester
Winter

Description
Waterloo SOS STAT 230 Final Review Package Prepared by Arin Goswami Edited by a Whole Lot of People Spring 2011 STAT 230 Final Review Package Spring 2011 Table of Contents Important formulas ........................................................................................................................ 3 Chapter 8 – Discrete Multivariate Distributions.............................................................................. 5 Examples:..................................................................................................................................... 7 Chapter 9 – Continuous Distributions........................................................................................... 13 Extra Practice for Final................................................................................................................... 20 2 Spring 2011 STAT 230 Final Review Package Important formulas 1. – (r) n = =n(n-1)(n-2)…(n-r+1) 2. – ( ) 3. 4. 5. | 6. | | | 7. ∑ | 8. ∑ ∑ a. b. c. 9. ∑ 10. [ ] [ ] [ ] 11. 12. √ 3 STAT 230 Final Review Package Spring 2011 13. ∑ ∑ 14. | | 15. [ ] ∑ 16. [ ] [ ] [ ] 17. 18. If X and Y are independent, then Cov(X, Y) = 0 19. The correlation coefficient of X and Y is √ 20. 21. ∑ ∑ ∑ ( ) 22. If we have n identically distributed random variables, and a = 1 for all i = 1, …, n i (∑ ) ∑ ( ) 4 STAT 230 Final Review Package Spring 2011 Chapter 8 – Discrete Multivariate Distributions Definitions 1. The expected value of a function of discrete rv’s X and Y, g(X, Y) is: [ ] ∑ This can be extended beyond two variables X and Y. 2. Property of Multivariate Expectation: [ ] [ ] [ ] 3. Variance: Some forms are easier to compute than others. [ ] ( ) [ ] [ ] 4. The covariance of X and Y, denoted [( )( )] Note: A handier formula for covariance is 5. If X and Y are independent, then Cov(X, Y) = 0. The converse is not true! 6. Suppose X and Y are independent random variables. Then, if and are any two functions, [ ] [ ] [ ]. 7. The correlation coefficient of X and Y is √ Note: This is a measure of the strength of the relationship between X and Y. lies in the interval [-1,1]. 8. Properties of Covariance: a. b. Intuition: Think of this as multiplying the two terms (aX+bY) and (cU+dV) together. (Which is exactly how it is derived using the definition) 9. Variance of a linear combination: 5 STAT 230 Final Review Package Spring 2011 In fact, more generally if we have n r.v’s X ,X ,…,X 1 2 n (∑ ) ∑ ∑ If we have n identically distributed random variables, and a = 1 for all I = 1, …, n i (∑ ) ∑ ( ) Note: This general formula is very useful in problems involving indicator random variables, which can only take on values 0 and 1. If all n random variables are independent, then (∑ ) ∑ 6 STAT 230 Final Review Package Spring 2011 Examples: Example 1 X and Y are two random variables that take on integer values from 0 to 2. You are given the following information about the distribution of X and Y: | (a) What is the correlation between X and Y? (b) let U = 5X and V = 3X – 2Y . Calculate Cov(U, V) Solutions (a) We want to draw a table for the joint distribution of X and Y. But first, note that | and we set P(X=0) = p giving P(X=0, Y=1)= 0.4p Using the given information, we fill in the following table Y 0 1 2 sum 0 0.12 0.4p 0.6p-0.12 P X 1 0.27-0.4p 0.23 0.4p 0.5 2 0.2 0.13-0.4p 0.17-0.6p 0.5-p Sum 0.59-0.4p 0.36 0.4p+0.05 1 Note: The bolded terms were calculated by first using the fact that and then by simple subtractions and additions. Make sure that you can replicate this table. 7 STAT 230 Final Review Package Spring 2011 Filling in the table for p=0.25, we have the following Y 0 1 2 Sum X 0 0.12 0.1 0.03 0.25 1 0.17 0.23 0.1 0.5 2 0.2 0.03 0.02 0.25 Sum 0.49 0.36 0.15 1 Now, recall that the correlation between X and Y is given by: √ From the table, we can calculate Var(X) and Var(Y) We were given that . Now, And, √ (b) Cov(U, V) = Cov(5X, 3X-2Y )=15Cov(X, X) – 10 Cov(X, Y ) Cov(X, X)= Var(X)= 0.5 2 2 2 Cov(X, Y )=E(XY )-E(X)E(Y ) ∑ 8 STAT 230 Final Review Package Spring 2011 Recall from part a) E(X) =1, E(Y )=0.96 Therefore, Cov(U, V) = Cov(5X, 3X-2Y )=15Cov(X, X) – 10 Cov(X, Y )= 15(0.5) – 10(-0.11) = 8.6 Example 2 The proportions of cats with blood types A, B, and AB in a large population are 0.7, 0.2 and 0.1 respectively. Let denote the frequencies of these three types in a random sample of size 20 taken from the population. Find the conditional distribution ogiven =12. For = 0,1,…,8 P( | = = = ( ) This is a binomial probability function. This makes sense since there are 8 trials where each trial will return blood type AB (Success) or type B (Failure). Given that we get either type AB or B, the probability of getting type AB on any given trail is 0.1/(0.1+0.2) = 1/3 9 Spring 2011 STAT 230 Final Review Package Example 3 1. Assume random variables X and Y have joint probability function as follows. x f(x,y) 0 1 2 0 0.2 0.3 0 y 2 0.05 0.2 0.25 a. Find the marginal probability function of X. f(x) = 0.25, 0.5, 0.25 for x = 0, 1, 2 b. Find Cov(X,Y) E(X) =1, E(Y) = 1, E(XY) = 2*0.2 + 4*0.25 = 1.4 So Cov (X, Y) = 1.4 – 1 = 0.4 c. Are X and Y independent? Why or why not? They are not independent—one justification is that covariance is non-zero. *Note: a covariance of 0 does not imply that they are independent! 10 STAT 230 Final Review Package Spring 2011 Example 4 Suppose that a pond contains 100 fish, and 40 of them are salmons. One day, 30 random fish are caught from the pond. Let X be the number of salmons caught. What is E(X) and Var(X)? Use indicator random variables to solve this problem. Solution: We first define indicator varia1le2 X ,30 ,… ,X , where { Also note that ∑ Justification: Forty out of 100 fishes in the pond are salmons. Now, (∑ ) ∑ ∑ Also, (∑ ) ∑ ∑ ∑ ( ) Now, ( ) And, ( ) ( ) ( ) Now, note that { ( ) Justification: For the fish i, we have a total of 100 fishes and 40 salmons. If the first fish is a salmon, then we have a total of 99 fishes and 39 salmons left. Thus E(i j = . This gives ( ) Hence, 11 STAT 230 Final Review Package Spring 2011 ( ) *Note that hypergeometric distribution would work for this question too 12 STAT 230 Final Review Package Spring 2011 Chapter 9 – Continuous Distributions Note: In this chapter, we present relevant examples after each group of definitions as there is a lot of material to cover. Definitions 1. The probability density function (p.d.f.) f(x) for a continuous random variable X is the derivative: Where F(x) is the c.d.f. for X. 2. The following are properties of a p.d.f. : a. ∫ b. c. ∫ ∫ d. ∫ 3. When X is continuous, we define [ ] ∫ Example 1 { be a pdf Find : a) k b) F(x) c) P(1/4 < X < 5/4) d) Var(X) Solutions: 13 STAT 230 Final Review Package Spring 2011 When finding the area of a region bounded by different functions, we split the integral into pieces: ∫ ∫ ∫ [ ] [ ] ( ) b) We start with the easy part, which we forget
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